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Let a planetary system be solely consisted of the earth and the sun with the earth revolving around the sun in a perfectly circular orbit. If the radial distance (R) of the earth from the sun is doubled, what will be the new revolving time (T)? If I apply Kepler's law of $T^2$ proportional to $R^3$ that yields T=1032 days. However, if I apply the conservation law of angular momentum:

Mass(m)$\times$angular speed(w)$\times$$R^2$= Constant,

or $w$$\times$$R^2$=constant,
or (2$\times$$\frac \pi T$)$\times$$R^2$=constant,
or $R^2$=constant$\times$T,
or T proportional to $R^2$; that is different from the Kepler's law! Can you please tell me in which assumption I am making the mistake?

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Angular momentum is conserved between two points on the same trajectory. In other words, the angular momentum of the Earth about the sun is the same right now as it was three hours ago.

This does not mean that the angular momentum of the Earth would be the same no matter how large its orbit is.

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  • $\begingroup$ Quite right. The conservation of angular momentum implies Kepler's 2nd law, the one that says that a planet (on a given trajectory!) sweeps out equal areas in equal times. $\endgroup$ – Bert Barrois Mar 26 '18 at 11:24
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Conservation of angular momentum applies to isolated systems. You cannot change the radius of circular orbit in your scenario, and retain the masses, without outside influence.

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  • $\begingroup$ dear, you have said that trajectory must be the same to apply the conservation law! However, in case of a rope revolving over the head with one side attached to a mass, how to calculate the revolving time if the person handling the rope suddenly doubles the rope? $\endgroup$ – Aziz Sartaz Chhoton Mar 26 '18 at 6:07
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In the Kepler Law $T^2 =kR^3$. When you increase the radius the constant does not change. K will have the same value i.e $ \frac{4\pi ^2}{GM} $ when you change the radius as it has no dependence on R.

The first equation you use is Angular momentum(MVR) = constant.
And The last equation that you put up is $R^2$ = constant $\times$ T.
When you increase the radius the angular momentum increases according to the first equation. And the constant changes. That means in the last equation as you change R to 2R, the constant value also changes and that means you'll have to take that under consideration as well.

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Your error is that you cannot assume a conservation of angular momentum when you double the orbital radius of earth. Keplers 3rd Law (for circular orbits) $$T_1^2:T_2^2=R_1^3:R_2^2 \tag1$$ follows from equating the centripetal force on earth with the gravitational force $$F_c=mR\omega^2=G\frac {m·M}{4\pi R^2}\tag 2$$ from which follow $$\omega^2·R^3=(\frac {2\pi}{T})^2·R^3=G\frac {M}{4\pi}=const \tag 3$$ which corresponds to Keplers's Law (1). The angular momentum of earth is given by $$L=m\omega R^2$$ Thus with eq. (3) this gives $$(\frac {L}{m})^2=const·R$$ This shows that the angular momentum of earth is proportional to the square root of the orbital radius $R$. When you double the orbital radius of earth, the angular momentum will not be constant but it will increase by a factor of $\sqrt{2}$.

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Well you have used conservation of momentum to get angular velocity, the time that u got there was the time for the planet to rotate and not revolve but the time u get when u use Keplers law of planetary motion is the time of revolution and not rotation hence the time periods are different.

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protected by Qmechanic Mar 29 '18 at 6:14

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